a: (x+2)(x-3)>0

nên x+2;x-3 cùng dấu

=>x>3 hoặc x<-2

b: (x-1)(x+4)<=0

nên x-1 và x+4 khác dấu

=>-4<=x<=1

\(=x^2-1+x-1-x^2-x+x^2-x+2x-2=-x^2\)

\(=-4+x^2+x=-x^2\)

\(=-4+x^2+x+x^2=0\)

\(=-4+2x^2+x=0\Rightarrow x=-1,687\)

a) x(x - 5) - 4x + 20 = 0

\(\Leftrightarrow\) x(x - 5) - (4x + 20)

\(\Leftrightarrow\) x(x - 5) - 4(x - 5) = 0

\(\Leftrightarrow\) (x - 5)(x - 4)

Khi x - 5 = 0 hoặc x - 4 = 0

 \(\Leftrightarrow\) x = 5           \(\Leftrightarrow\) x = 4

 Vậy S = \(\left\{5;4\right\}\)

b) x(x + 6) - 7x - 42 = 0

 \(\Leftrightarrow\) x(x + 6) - (7x - 42) = 0

 \(\Leftrightarrow\) x(x + 6) - 7(x + 6) = 0

 \(\Leftrightarrow\) (x + 6)(x - 7) = 0

Khi x - 6 = 0 hoặc x - 7 = 0

   \(\Leftrightarrow\) x = 6           \(\Leftrightarrow\) x = 7

 Vậy S = \(\left\{6;7\right\}\)

c) x³ - 5x² - x + 5 = 0

 \(\Leftrightarrow\) (x³ - 5x²) - (x + 5) = 0

 \(\Leftrightarrow\) x² (x - 5) - (x - 5) = 0

 \(\Leftrightarrow\) (x - 5)(x² - 1) = 0

 \(\Leftrightarrow\) (x - 5)(x - 1)(x + 1) = 0

 Khi x - 5 = 0 hoặc x - 1 = 0 hoặc x + 1 = 0

   \(\Leftrightarrow\) x = 5           \(\Leftrightarrow\) x = 1            \(\Leftrightarrow\) x = -1

 Vậy S = \(\left\{5;1;-1\right\}\)

d) 4x² - 25 - (2x - 5)(3x + 7) = 0

\(\Leftrightarrow\) (2x)² - 5² - (2x - 5)(3x + 7) = 0

\(\Leftrightarrow\) (2x - 5)(2x + 5) - (2x - 5)(3x + 7) = 0

\(\Leftrightarrow\) (2x - 5) \([\left(2x+5\right)-\left(3x+7\right)]\) = 0

\(\Leftrightarrow\) (2x - 5) ( 2x + 5 - 3x + 7) = 0

\(\Leftrightarrow\) (2x - 5)( -x + 12) = 0

Khi 2x - 5 = 0 hoặc -x + 12 = 0

  \(\Leftrightarrow\) 2x = 5             \(\Leftrightarrow\)   -x = -12

  \(\Leftrightarrow\) x = \(\dfrac{5}{2}\)              \(\Leftrightarrow\) x = 12

 Vậy S = \(\left\{\dfrac{5}{2};12\right\}\)

e) x³ + 27 + (x + 3)(x - 9) = 0

\(\Leftrightarrow\) x³ - 3³ + (x + 3)(x - 9) = 0

\(\Leftrightarrow\) (x - 3)(x² - 3x + 9) + (x + 3)(x - 9) = 0

\(\Leftrightarrow\) (x - 3) \(\left[\left(x^2-3x+9\right)+\left(x-9\right)\right]\) = 0

\(\Leftrightarrow\) (x - 3) ( x² - 3x + 9 + x - 9) = 0

\(\Leftrightarrow\) (x - 3)(x² - 2x) = 0

\(\Leftrightarrow\) (x - 3)x(x - 2)

 Khi x - 3 = 0 hoặc x = 0 hoặc x - 2 = 0

    \(\Leftrightarrow\) x = 3                            \(\Leftrightarrow\) x = 2

 Vậy S = \(\left\{3;0;2\right\}\)

 Chúc bạn học tốt

a) Ta có: \(x\left(x-5\right)-4x+20=0\)

\(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\end{matrix}\right.\)

b) Ta có: \(x\left(x+6\right)-7x-42=0\)

\(\Leftrightarrow x\left(x+6\right)-7\left(x+6\right)=0\)

\(\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)

a: \(x^4-2x^3-25x^2+50x=0\)

=>\(x^3\left(x-2\right)-25x\left(x-2\right)=0\)

=>\(\left(x-2\right)\left(x^3-25x\right)=0\)

=>x(x-2)(x^2-25)=0

=>x(x-2)(x+5)(x-5)=0

=>x∈{0;2;-5;5}

b: \(x^2\left(x-1\right)-4x^2+8x-4=0\)

=>\(x^2\left(x-1\right)-4\left(x^2-2x+1\right)=0\)

=>\(\left(x-1\right)\left(x^2-4x+4\right)=0\)

=>\(\left(x-1\right)\left(x-2\right)^2=0\)

=>x∈{1;2}

c: \(9x^2-4-2\left(3x-2\right)^2=0\)

=>(3x-2)(3x+2)-(3x-2)(6x-4)=0

=>(3x-2)(3x+2-6x+4)=0

=>(3x-2)(-3x+6)=0

=>(x-2)(3x-2)=0

=>x∈{2;2/3}

d: \(9x^2+90x+225-\left(x-7\right)^2=0\)

=>\(\left(3x+15\right)^2-\left(x-7\right)^2=0\)

=>(3x+15-x+7)(3x+15+x-7)=0

=>(2x+22)(4x+8)=0

=>2(x+11)*4*(x+2)=0

=>(x+11)(x+2)=0

=>x∈{-11;-2}

e: \(x^3-8+\left(x-2\right)\left(x+1\right)=0\)

=>\(\left(x-2\right)\left(x^2+2x+4\right)+\left(x-2\right)\left(x+1\right)=0\)

=>\(\left(x-2\right)\left(x^2+2x+4+x+1\right)=0\)

=>\(\left(x-2\right)\left(x^2+3x+5\right)=0\)

mà \(x^2+3x+5=x^2+3x+\frac94+\frac{11}{4}=\left(x+\frac32\right)^2+\frac{11}{4}>0\forall x\)

nên x-2=0

=>x=2

g: (x+1)(x+2)(x+3)(x+4)-24=0

=>\(\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=0\)

=>\(\left(x^2+5x\right)^2+10\left(x^2+5x\right)+24-24=0\)

=>\(\left(x^2+5x\right)\left(x^2+5x+10\right)=0\)

mà \(x^2+5x+10=x^2+5x+\frac{25}{4}+\frac{75}{4}=\left(x+\frac52\right)^2+\frac{75}{4}\ge\frac{75}{4}>0\forall x\)

nên \(x^2+5x=0\)

=>x(x+5)=0

=>x∈{0;-5}

mk cảm ơn ah!!!

5 x 5 = 25 

Học tốt

Nhớ t.i.c.k

#Vii

5 × 5 = 25

Chúc bạn có một kì nghỉ hè thật vui vẻ

\(ac=-6< 0\Rightarrow\) phương trình đã cho luôn luôn có 2 nghiệm pb (trái dấu)

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=m-2\\x_1x_2=-6\end{matrix}\right.\)

Thế vào đề bài:

\(m-2-3\left(-6\right)=0\)

\(\Leftrightarrow m+16=0\Leftrightarrow m=-16\)

Thầy phân tích cho e kĩ hơn ở p [ac=-6] đc ko ạ. Tại sao mk ko tính Δ= [m^2-4m+28 kết quả tính đc] mà p làm như thế ạ

a, \(2x\left(x-3\right)-15+5x=0\\ \Rightarrow2x\left(x-3\right)-\left(15-5x\right)=0\\ \Rightarrow2x\left(x-3\right)-5\left(3-x\right)=0\\ \Rightarrow\left(2x+5\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{2}\\x=3\end{matrix}\right.\)

b, \(x^3-7x=0\\ \Rightarrow x\left(x^2-7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\pm7\end{matrix}\right.\)

c, \(\left(2x-3\right)^2-\left(x+5\right)^2=0\\ \Rightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\\ \Rightarrow\left(x-8\right)\left(3x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-\dfrac{2}{3}\end{matrix}\right.\)

Xem lại đề câu d 

\(\dfrac{x+2y}{4x-3y}=-2\)

=>x+2y=-8x+6y

=>9x=4y

hay x/y=4/9